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QUESTION


\begin{description}

\item[(a)]
Let $(G,*)$ be a group. Carefully prove the following statement
using only the axioms for a group, indicating which axiom you used
at each stage of the argument: There is a unique element $h\in G$
such that $h*g=g$ for every element $g\in G$.

\item[(b)]
State Lagrange's Theorem and use it to prove that if $p$ and $q$
are prime and $G$ is a group of order $pq$ then every proper
subgroup $G$ is cyclic. (A proper subgroup is one not equal to
$G$.)

\item[(c)]
Write out the Cayley table for the group of symmetries of an
equilateral triangle using the notation $s$ to represent the
anticlockwise rotation through $\pi/3$ and $x,y,z$ to denote the
three reflections in the lines $X,Y,Z$ as marked in figure 1.

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\put(3,1){\line(-2,3){1}}

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\put(1.3,1.9){$X$} \put(.7,1){$Y$} \put(1.7,.7){$Z$}

\put(1.5,.3){Figure 1}

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\item[(d)]
For each of the statements below either show it is true OR give an
example to show that it is false:

\begin{description}

\item[(i)]
If $g$ and $h$ are elements of a group and both have order $n$
then their product also has order $n$.

\item[(ii)]
If every proper subgroup of a group $G$ is cyclic then so is $G$.

\item[(iii)]
Every group of order 8 contains a cyclic subgroup of order 4.

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ANSWER


\begin{description}

\item[(a)]
By the identity axiom there is an element $e\in G$ such that
$e*g=g\forall g\in G$. Now suppose $h\in G$ and $h*g=g\forall g\in
G$. In particular $h*h=h$.

By the inverse axiom there is an element $h^{-1}\in G$ such that
$h^{-1}*h=e$ so $h^{-1}*(h*h)=h^{-1}*h=e$.

By associativity $(h^{-1}*h)*h=h^{-1}*(h*h)$ so
$e*h=(h^{-1}*h)*h=h^{-1}*(h*h)=e$ or $h=e$.

\item[(b)]
Lagrange's Theorem

If $G$ is a finite group and $H$ is a subgroup of $G$ then $|H|$
divides $|G|$.

If $|G|=pq$ with $p,q$ prime then any proper subgroup $h<G$ has
$|H|=1,p$ or $q$.

Since $p,q$ are prime $H$ is cyclic.

\item[(c)]

\begin{tabular}{c|cccccc}
$\circ$&$e$&$s$&$s^2$&$x$&$y$&$z$\\ \hline
$e$&$e$&$s$&$s^2$&$x$&$y$&$z$\\ $s$&$s$&$s^2$&$e$&$y$&$z$&$x$\\
$s^2$&$s^2$&$e$&$s$&$z$&$x$&$y$\\ $x$&$x$&$z$&$y$&$e$&$s^2$&$s$\\
$y$&$y$&$x$&$z$&$s$&$e$&$s^2$\\ $z$&$z$&$y$&$x$&$s^2$&$s$&$e$
\end{tabular}

OR

\begin{tabular}{c|cccccc}
$\circ$&$e$&$s$&$s^2$&$x$&$y$&$z$\\ \hline
$e$&$e$&$s$&$s^2$&$x$&$y$&$z$\\ $s$&$s$&$s^2$&$e$&$z$&$x$&$y$\\
$s^2$&$s^2$&$e$&$s$&$y$&$z$&$x$\\ $x$&$x$&$z$&$y$&$e$&$s$&$s^2$\\
$y$&$y$&$z$&$x$&$s^2$&$e$&$s$\\ $z$&$z$&$x$&$y$&$s$&$s^2$&$e$
\end{tabular}


\item[(d)]

\begin{description}

\item[(i)]
False, $x,y\in D_3$ above have order 2, $xy=s^2$ has order 3.

\item[(ii)]
False, Every proper subgroup of $D_3$ is cyclic but $D_3$ is not.

\item[(iii)]
False $C_2\times C_2 \times C_2$

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